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Basically the same as Formatting Sandbox in Meta Stack Exchange, but since this and Statistical Analysis are the only two sites (I know) supporting $\TeX$ formatting, I believe we also need one here for testing it.

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    $\begingroup$ Theoretical computer science also supports $\mathrm{\TeX/\LaTeX}$ formatting. $\endgroup$ – JeffE Jun 1 '12 at 7:29
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    $\begingroup$ @JeffE: You can use $\TeX$ and $\LaTeX$ (\Tex and \LaTeX) for the text. $\endgroup$ – Asaf Karagila Jun 2 '12 at 21:09
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    $\begingroup$ @JeffE: In 2010 only 'stats' and 'math' support TeX formatting. Of course now there is also 'cstheory', 'cs', 'chemistry', 'quant', etc. $\endgroup$ – kennytm Jun 3 '12 at 6:25
  • $\begingroup$ test $$\begin{align*}\text{middle line}\end{align*}$$ new line $\endgroup$ – Ruslan Jan 30 '14 at 13:06
  • $\begingroup$ test test $\not\in(1)\notin(2)$ Who's better??? $\endgroup$ – user93957 Jan 31 '14 at 22:25
  • $\begingroup$ $m^n + m^x + m^n = 555555$ test test $\endgroup$ – hichris123 Feb 2 '14 at 19:09
  • $\begingroup$ line $\begin{array}\phantom{i}\\\phantom{i} \end{array} $ line2 $\endgroup$ – Bill Dubuque Apr 18 '14 at 15:15
  • $\begingroup$ quotes test ${``}=\text{’’}$ $\endgroup$ – Ruslan Oct 25 '17 at 11:13
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    $\begingroup$ $a``=\!\!\text{’’}b$ $\endgroup$ – Ruslan Oct 25 '17 at 11:17
  • $\begingroup$ I would understand the newly covered elements at the time we choose $S_3$ should be $S_3\setminus(S_1\cup S_2)$ instead of $S_3\setminus(S_1\cap S_2\cap S_3)$. $\endgroup$ – Apass.Jack Aug 25 '18 at 20:51

43 Answers 43

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Demonstrating the issue from this question

$\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}$

$\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b} + \frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b} - \frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b} = \frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\times\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}\frac{a}{b}$

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Sudoku test $$\begin{array}{||c|c|c||c|c|c||c|c|c||} \hline\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 \\ \hline 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\hline 2 & 3 & 1 & 5 & 6 & 4 & 8 & 9 & 7 \\ \hline 5 & 6 & 4 & 8 & 9 & 7 & 2 & 3 & 1 \\ \hline 8 & 9 & 7 & 2 & 3 & 1 & 5 & 6 & 4 \\ \hline\hline 3 & 1 & 2 & 6 & 4 & 5 & 9 & 7 & 8 \\ \hline 6 & 4 & 5 & 9 & 7 & 8 & 3 & 1 & 2 \\ \hline 9 & 7 & 8 & 3 & 1 & 2 & 6 & 4 & 5 \\ \hline\hline \end{array}$$

Result: Doesn't work as expected. Actual $\rm\LaTeX$ output of the same code:

LaTeX output

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Assuming $0<a<b<c,$ $$\int_0^\infty\operatorname{arccot}(ax)\,\operatorname{arccot}(bx)\,\operatorname{arccot}(cx)\,dx=\frac{I_{\text{log}}+I_{\text{dilog}}+I_{\text{trilog}}}{24abc} $$


$$ I_{\text{log}}=\pi ^2 \left[\vphantom{\Large|}a c \left\{\vphantom{\large|}\ln (c-b)+5 \ln (b+c)-6 \ln c\right\}\\ +b \left\{\vphantom{\large|}4 (a+c) \ln (a+c)-(2 a+c) \ln (c-b)+(4 a+3 c) \ln (b+c)-6 (a \ln b+c \ln c)\right\}\right]\\ +a c \left[\ln ^3\left(\frac{(a+b)c}{(b+c)a}\right)-\ln ^3\left(\frac{(a+b)c}{(c-b)a}\right)\\ -6 \ln \left(\frac a c\right) \left\{\ln \left(\frac{(a+b) c}{(c-b)a}\right)\cdot \ln \left(\frac{(a+c)b}{(c-b)a}\right)+\ln \left(\frac{(c-a)b}{(b+c)a}\right)\cdot \ln \left(\frac{(b+c)a}{(a+b) c}\right)\right\}\right]\\ +a b \left[3 \ln \left(\left(b^2-a^2\right) c\right)\cdot \ln ^2(a+c)-2 \ln ^3(a+c)-6 \ln (a+b)\cdot \ln (c-b)\cdot \ln (a+c)\\ -6 \ln ((b-a) c)\cdot \ln (b+c)\cdot \ln (a+c)-2 \ln ^3(c-b)-2 \ln ^3(b+c)\\ +3 \ln ((a+b) (a+c)c)\cdot \ln ^2(c-b)+3 \ln ((b-a) (a+c)c)\cdot \ln ^2(b+c)\\ +3 \ln c\cdot \ln (c-a)\cdot \ln \left(\frac{c-a}{(c-b)^2}\right)-3 \ln b\cdot \left\{\ln ^2\left(\frac{a-c}{b-c}\right)+\ln ^2\left(\frac{a+c}{b+c}\right)\right\}\\ +\ln^2\left(\frac a b\right)\cdot \left\{\vphantom{\Large|}8 \ln a-2 \ln b-6 \ln ((b-a) c)-3 \ln \left(c^2-a^2\right)+6 \ln \left(c^2-b^2\right)\right\}\right]\\ +b c \left[\left(3 \ln \left(b^2-a^2\right)-2 \ln (a+c)\right) \ln ^2(a+c)+3 \ln \left(\frac{a \left(b^2-a^2\right) (a+c)}{b+c}\right) \ln ^2(b+c)\\ +\ln (c-b) \left(-6 \ln (b) \ln (c-a)+6 \ln \left(\frac{a}{a+b}\right) \ln (a+c)-\ln (c-b) \ln \left(\frac{a^3 (c-b)}{(a+c)^3}\right)\right)\\ -6 \left(\ln (a) \ln (c-a)+\ln \left(1-\frac{a}{b}\right) \ln (a+c)\right) \ln (b+c)\\ -3 \ln\left(\frac{c+a}{c-a}\right) \left(\ln ^2(b)-\ln (a) \ln \left(\frac{a b^2}{(a+b)^2}\right)+\ln \left(\frac{b}{c}\right) \ln \left(c^2-a^2\right)\\ -2 \ln (c) \ln \left(\frac{(a+b) c}{a^2 \left(c^2-b^2\right)}\right)\right)\\ +6 \ln\left(\frac{c+b}{c-b}\right) \left(\ln ^2\left(\frac{a}{b}\right)-\ln ^2(a+b)+\ln ^2\left(\frac{b}{c}\right)-\ln (b) \ln \left(c^2-b^2\right)\right)\right] $$


$$ I_{\text{dilog}}=6 \left(a b \ln \left(\frac{a}{b}\right)-a c \ln \left(\frac{a}{c}\right)+b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{(b-a) c}{a (b-c)}\right)-6 \left(-a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)+b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{b (a+c)}{a (b-c)}\right)+6 \left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)+b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{b (a-c)}{a (b+c)}\right)+6 \left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)-b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{(a-b) c}{a (b+c)}\right)$$


$$I_{\text{trilog}}=12 a b \left(\operatorname{Li}_3\left(-\frac{a}{b}\right)-\operatorname{Li}_3\left(\frac{a}{b}\right)\right)+12 a c \left(\operatorname{Li}_3\left(-\frac{a}{c}\right)-\operatorname{Li}_3\left(\frac{a}{c}\right)\right)+12 b c \left(\operatorname{Li}_3\left(-\frac{b}{c}\right)-\operatorname{Li}_3\left(\frac{b}{c}\right)\right)-6 a (b-c) \left(\operatorname{Li}_3\left(\frac{b-a}{b-c}\right)-\operatorname{Li}_3\left(\frac{a+b}{b-c}\right)-\operatorname{Li}_3\left(\frac{(b-a) c}{a (b-c)}\right)\right)-6 (a-b) c \operatorname{Li}_3\left(\frac{b-a}{b+c}\right)+6 (a+b) c \left(\operatorname{Li}_3\left(\frac{a (c-b)}{(a+b) c}\right)+\operatorname{Li}_3\left(\frac{a+b}{b+c}\right)\right)+6 a (b+c) \operatorname{Li}_3\left(\frac{(a-b) c}{a (b+c)}\right)-6 b (a-c) \left(\operatorname{Li}_3\left(\frac{a-b}{a-c}\right)-\operatorname{Li}_3\left(\frac{(a-b) c}{b (a-c)}\right)-\operatorname{Li}_3\left(\frac{b (a-c)}{a (b+c)}\right)+\operatorname{Li}_3\left(\frac{c-a}{b+c}\right)\right)-6 (a+b) c \operatorname{Li}_3\left(\frac{a (b+c)}{(a+b) c}\right)+6 b (a+c) \left(\operatorname{Li}_3\left(\frac{a+b}{a+c}\right)+\operatorname{Li}_3\left(\frac{(b-a) c}{b (a+c)}\right)+\operatorname{Li}_3\left(\frac{b (a+c)}{a (b-c)}\right)+\operatorname{Li}_3\left(\frac{a+c}{b+c}\right)-2 \zeta (3)\right).$$

Mathematica expression is here.

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Free to use! Have fun!kfdj;klfjasdkl

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Replace an SEO unfriendly pic question

a sample pic question

with a markdown blockquote for site SEO (in order to avoid harming the content originality by enclosing cited contents with <blockquote> using markdown)

The following sum $$ \sqrt{25-\left(\frac5n\right)^2} \cdot \frac5n + \sqrt{25-\left(\frac{10}n\right)^2} \cdot \frac5n + \cdots + \sqrt{25-\left(\frac{5n}{n} \right)^2} \cdot \frac5n $$ is a right Riemann sum for the definite integral $$\int_a^b f(x)\,\mathrm{d}x$$ where $b = \:\bbox[2px, white, border:1px solid black]{\color{white}{\Rule{5em}{1em}{0.1em}} {\tt 5}}$
and $f(x) = \:\bbox[2px, white, border:1px solid black]{\color{white}{\Rule{1em}{1em}{0.1em}} {\tt sqrt(25 - x^2)}}$.
The limit of these Riemann sums as $n \to \infty$ is  $\bbox[2px, white, border:1px solid black]{\color{white}{\Rule{8em}{1em}{0.1em}}}$ .

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Aligning equations: \begin{align} 3x-1 &= -10 \\ 3x &= -9 \\ x &= -3 \end{align}

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This post has been idling for a while. Sorry Steve.

Test for trickeries on hyperlinks.

The main objective is to get the tag (visual cue) on the left, at the beginning of a line, be it in mathmode or text (sectioning).

Does this work? Yes, and inline \ref{sec01} tag works \ref{sec02} just as well in a standard mode or \ref{keypoint}. Now the challenge is to $\hspace{600pt}\ref{sec03}$. Hell yeah!

So let's see how these silly ones work. First there's the blank Eq.\eqref{05} and the 2nd Eq.\eqref{12}, both tags consisting of 3 white spaces. Then we have Eq.\eqref{13} and lastly Eq.\eqref{14}. oh how about not in eqref but jus ref like these Eq.\ref{05}, Eq.\ref{12}, Eq.\ref{13}, and Eq.\ref{14}?

jaydfox I've already touched on this, but I did want to describe somewhat briefly how I've accelerated convergence of Andrew's slog, with a more detailed description to follow when I have the time.

\begin{align*} &\text{This is an} & &\text{align environment} & &\text{Seriously} \tag{ } \label{05} \\ \mathbb{E}[X] &\equiv \mathcal{X} & \frac{\mathscr{X}}{\mathbb{Y}} &\approx \frac45 \Theta & j &= 2 \tag*{ } \label{12}\\ A &\equiv B+C & \frac{x}y &\approx \frac12 \Theta & i &= 3 \tag*{no kidding} \label{13}\\ \tag*{tag and label in 1st column} \label{14} \alpha &\equiv \beta+\gamma & \frac{\eta}\xi &\approx \frac23 \Omega & k &= 4 \\ \end{align*}

First of all, let's make the assumption that the solution to the infinite system exists, and furthermore, that the resultant power series has a nonzero radius of convergence. Although heuristically the evidence is fairly strong, we don't know enough to formally prove it. However, like the Riemann hypothesis, we can still make some solid conclusions which we know are true IF the hypothesis is true.

So this is a hyperlink to Eq.\,\eqref{01} which is supposed to be the normal tag, and then Eq.\,\eqref{02} is supposed to clash with Eq.\,\eqref{03}. Of course, the standard use is like Eq.\,\eqref{04}. oh damn it no thin space unless in mathmode so maybe just Eq.\eqref{04} unless one goes through the trouble of $\text{Eq.}\,\eqref{04}$? Yeah there IS a visible diffrence.

Anyway, based on these assumptions, let's try to figure out how the convergence of the smaller systems works. First, let's start with the infinite system.
We'll call the matrix M, and the column vector $[1, 0, 0, ...]$ we'll call B. The coefficients A of the slog() are then the solution of the equation MA = B.

Now this is getting interesting: \Eq.\eqref{21}, or should I call \Eq.\eqref{22}? Getting into the realm of absurdity with \Eq.\eqref{23} and oh btw \Eq.\eqref{24} also with negative that's gonna mess things up, and \Eq.\eqref{25} is no exception. With just \ \ref we have \Eq.\ref{21} ipsum oh I cannot remember \Eq.\ref{22} yes lorem ipsum it is \Eq.\ref{23} can I block call this thing \Eq.\ref{24} by \require \Eq.\ref{25}?

Notice that we don't actually calculate the solution to the full system. We end up with, for example, a $100 \times \infty$ system. $100$ rows, infinity columns. However, in practice, we truncate this down to perhaps $100 \times 2000$. And we don't even have to solve the full system. We can precalculate the $100 \times 1900$ system, starting at column $101$, and subtract from B .

Now, below a certain row, say, row $100$, let's replace the rows with the corresponding entry from an infinite identity matrix. So rows $1..100$ will stay the same, but $101 .. \infty$ are going to be changed.
In order to do this, we replace the zero in the column vector B with the actual coefficient in A we're trying to solve for. This requires knowing the coefficients in A, obviously.

$$s(m-1,n)= 2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}}(-1)^{\frac{n}{2}-i}{m+n+2i-2 \choose 2i}{n\choose \frac{n}{2}-i} \tag{clash tag} \label{02} $$

$$s(m,n)= 2^{m-2}\sum\limits_{i=0}^{\frac{n}{2}}(-1)^{\frac{n}{2}-i}{m+n+2i-1 \choose 2i}{n\choose \frac{n}{2}-i} \tag{normal tag} \label{01} $$

$$ s(m,n-1)=2^{m-2}\sum\limits_{i=0}^{\frac{n-2}{2}}(-1)^{\frac{n-2}{2}-i}{m+n+2i-1 \choose 2i}{n-1\choose \frac{n-2}{2}-i} \tag{clash tag} \label{03} $$

$$ s(m-1,n-1)=2^{m-3}\sum\limits_{i=0}^{\frac{n-2}{2}}(-1)^{\frac{n-2}{2}-i}{m+n+2i-2 \choose 2i}{n-1\choose \frac{n-2}{2}-i} \tag{4} \label{04} $$ Now it should be clear how Andrew's slog is converging on the true solution. We're solving for the infinite system, using the modified format the I described above, but instead of replacing the values in B with the true values, we're replacing them with $0$'s. This introduces errors in the first $100$ (or n) coefficients to compensate.

Extending further, I can solve a $700 \times 700$ system, using a precomputed vector which gives me the solution to a $700 \times 10,000$ system, which itself is the solution to a $10,000 \times 10,000$ system with approximations for terms $701$ to $10,000$. As with the original system, the errors stack up about halfway through, so a $700 \times 700$ system is only really accurate out to about $300-400$ terms. I should be careful about how I say this, because all $700$ terms are at least as accurate as the approximating series I would get if I used the coefficients of the two singularities.


$\Large\color{green}{\text{Oh my god I cannot believe this works!!}}$

Let's quote without the push \Eq.\eqref{31} and with the deliberate \Eq.$\hspace{100pt}\eqref{31}$ positive space paired with the equation labels \Eq.$\hspace{2000pt}\eqref{32}$, Oh my god I cannot believe this works. But is that the inevitable double parenthesis? How about \Eq.$\hspace{300pt}\eqref{33}$, is there still this thing? Now this \Eq.$\hspace{400pt}\eqref{34}$ and again \Eq.$\hspace{500pt}\eqref{35}$ and on and on to \Eq.$\hspace{600pt}\eqref{36}$. Hold on let's see $\hspace{100pt}\ref{31}$, $\hspace{200pt}\ref{32}$, $\hspace{300pt}\ref{33}$. Yeah, these show that $\hspace{400pt}\ref{34}$ tag position push $\hspace{500pt}\ref{35}$ is not affected by equation content, and we have $\hspace{600pt}\ref{36}$. Okay so there's no need to manually tag (label) the parenthesis just one-sided? Now do these $\hspace{700pt}\ref{37}$ go over the boarder $\hspace{800pt}\ref{38}$? Yes they totally do. So it is capped at 600? How about $\hspace{600pt}\ref{37}$ and $\hspace{600pt}\ref{38}$? Yes, exactly 600pt….. wait, there’s a HUGE problem to this system with any negative space: the page jumps to the middle (or below) the equation and cutting the letters in half, unlike the standard setup that jumps to show the entire equation regardless of the height of the expression. The immediate response to solve this is to always setup a phantom line just above $\tag*{wherever}\label{keypoint}$ you want to jump to (ha, not exactly like this), and maybe use some commands the squash the height of the phantom line. This is less than

$\Large\color{green}{\text{I guess for the sake of sectioning}\ldots} \tag*{Sec.1} \label{sec01}$

. . . it’s okay to setup a phantom line (the whole section heading is in mathmode anyway).


The residue is the part that is inaccurate, and by the $300$th term, the coefficients seem to be about ten orders of magnitude smaller than the singularities I've re-moved. And by the $300$th coefficient, I'm already dealing with extremely small coefficients as it is. So this $700 \times 700$ system is probably accurate to a hefty number of decimal places, at least $40$, if not $60$ or more, for complex inputs about a unit distance from the origin or less.

In order to improve convergence, what I've done is taken the coefficients for the two known singularities at $0.31813... + 1.3372357i$ and its conjugate, and put them in the B vector, rather than $0$'s. These coefficients aren't the true values from A, but they're much, much, much closer to the true values. This reduces the errors in the final solution by several orders of magnitude.

\begin{gather*} s(m-1,n-1)=2^{m-3}\sum\limits_{i=0}^{\frac{n-2}{2}}(-1)^{\frac{n-2}{2}-i}{m+n+2i-2 \choose 2i}{n-1\choose \frac{n-2}{2}-i} \tag{21} \label{21} \\ =2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-1-i}{m+n+2i-2 \choose 2i}{n-1\choose \frac{n}{2}-1-i} \tag{$\color{red}{red tag}$} \label{22} \\ =-2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-i}{m+n+2i-2 \choose 2i}\frac{(n-1)!}{(\frac{n}{2}-1-i)!(\frac{n}{2}+i)!} \tag{white $\color{white}{tag}$} \label{23} \\ =-2^{m-3}\sum\limits_{i=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-i}{m+n+2i-2 \choose 2i}\frac{n!}{(\frac{n}{2}-i)!(\frac{n}{2}+i)!}\frac{\frac{n}{2}-i}{n} \tag{$\hspace{-120pt}\color{blue}{\Large blue Large}$ tag negative space} \label{24}\\ n\equiv 0 \pmod 2 \tag{$\hspace{-120pt}\color{magenta}{\large magenta large}$ same -120pt} \label{25} \end{gather*}

In other words, assume we're given K as a $100 \times 2000$ matrix, A as a column vector with $2000$ rows, and B as a column vector with $100$ rows. Then, using the known values of K, and the approximations for rows $101..2000$ of A, we solve the new system:

$$ K_{ 1...100} A_{1...100}\ =\ B- K_{101...2000} A_{ 101...2000}$$

Thus, we only solve a $100 \times 100$ system, which actually is solving a $2000 \times 2000$ system, with very, very good approximations of the coefficients $101..2000$. And note that the right hand side can be precalculated in chunks, reducing memory requirements and making it feasible to solve large systems. For example:

\begin{gather*} D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-100pt}$(-100pt)} \label{31} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-200pt}$(-200pt)} \label{32} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-300pt}$(-300pt)} \label{33} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-400pt}$Eq.(-400pt)} \label{34} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-500pt}$Eq.(-500pt)} \label{35} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-600pt}$Eq.(-600pt)} \label{36} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-600pt}$Eq.(-700pt)} \label{37} \\ D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)). \tag*{$\hspace{-600pt}$Eq.(-800pt)} \label{38} \end{gather*} Lorem ipsum dolor sit amet, consectetur adipiscing elit. Vivamus bibendum a quam eu vulputate. Phasellus congue velit mi, non gravida neque lobortis ut. Quisque dictum erat eget neque imperdiet lobortis. Vivamus vel malesuada purus. Nunc vulputate sem sit amet pretium luctus. Duis interdum, magna id interdum eleifend, est nisl malesuada dolor, sed mollis metus metus ac magna. Ut nisl risus, tempor eu dolor vel, suscipit luctus elit. Pellentesque facilisis arcu vitae lacus rutrum viverra. Ut enim sem, imperdiet non volutpat at, ultricies id est. Aenean ex mauris, scelerisque id dolor nec, vehicula tincidunt ligula. Proin non egestas dui, in tincidunt ante. Nunc placerat, nunc ac varius molestie, ipsum ipsum molestie arcu, id rhoncus tellus nunc commodo sapien.

$\tag*{$\large\color{green}{\textit{Section.2}}$} \label{sec02}$

In tincidunt finibus porttitor. Integer lobortis dignissim iaculis. Mauris metus lectus, posuere eu cursus a, tempus sed nunc. Donec ut purus efficitur, pellentesque purus id, auctor erat. Curabitur accumsan sed nisi vitae finibus. Quisque vitae nisl eu eros egestas sollicitudin in vel turpis. Donec dapibus, nibh nec fringilla suscipit, turpis leo tempor nibh, eleifend accumsan nisi mauris quis eros. Nunc malesuada suscipit quam non viverra. Praesent ante justo, semper non efficitur non, ullamcorper eget nulla. Pellentesque quis mi sit amet nulla consequat iaculis. Morbi consectetur augue nec leo aliquet, at venenatis est malesuada. Sed dapibus vehicula velit a rhoncus. Morbi a mi sollicitudin, accumsan odio ut, viverra turpis.

$\tag*{$\hspace{-600pt}\large\color{blue}{\textbf{Section.3}}$} \label{sec03}$

Nullam vulputate fermentum vestibulum. Integer dictum libero et sem consectetur, id consequat odio vestibulum. Nullam vel efficitur mi. Phasellus turpis massa, volutpat id blandit vitae, lacinia vel quam. Proin vitae ex quis mauris blandit molestie fringilla vestibulum augue. Phasellus euismod dictum eros sit amet vulputate. Nam condimentum convallis nisi, id tristique libero porttitor sit amet. Praesent eleifend nisi sed eros bibendum, ut tincidunt nunc mattis. Pellentesque mi metus, ultrices congue tellus sit amet, semper bibendum ligula. In risus eros, tempor et tempus sit amet, imperdiet at dolor. Nam quis tristique elit. Praesent venenatis varius euismod. Fusce dignissim lorem sed odio dignissim, eu congue ex malesuada.

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Functions are generally $\mathcal{M}_{8\times 8}[\operatorname{GF}(2^8)]\rightarrow\mathcal{M}_{8\times 8}[\operatorname{GF}(2^8)]$ with input matrices $\in\mathcal{M}_{8\times 8}[\operatorname{GF}(2^8)]$.

$$\begin{align} \mu(a) &= b \Leftrightarrow b_{ij} = a_{8i + j}\text{, }0\leqslant i\text{, }j\leqslant 7\\ \gamma(a) &= b\Leftrightarrow b_{ij} = S[a_{ij}]\text{, }0\leqslant i\text{, }j\leqslant 7\\ \pi(a) &= b\Leftrightarrow b_{ij} = a_{(i-j)\bmod 8,j}\text{, }0\leqslant i\text{, }j\leqslant 7\\ \sigma[k](a) &= b \Leftrightarrow b_{ij} = a_{ij}\oplus k_{ij}\text{, }0\leqslant i\text{, }j\leqslant 7\\ \\ c^{r}_{0j} &\equiv 0\text{, }1\leqslant i\leqslant 7\text{, }0\leqslant j\leqslant 7\\ c^{r}_{0j} &\equiv S[8(r - 1) + j]\text{, }0\leqslant j\leqslant 7\\ \\ \rho[k] &\equiv \sigma[k]\circ\theta\circ\mu\circ\gamma\\ \\ K^0 &= K\\ K^r &= \rho[c^r](K^{r-1})\text{, } r \ge 0\\ \\ W[K] &= \biggl(\underset{1}{\overset{r=R}{\bigcirc}} \rho[K^r]\biggr)\circ \sigma[K^0]\\ \\ \eta_i &= \mu(m_i)\\ H_0 &= \mu(IV)\\ H_i &= W[H_{i-1}](\eta_i)\oplus H_{i-1}\oplus\eta_i\text{, }1\leqslant i\leqslant t\\ \\ \operatorname{Whirlpool}(M) &\equiv \mu^{-1}(H_t) \end{align}$$

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$$\begin{align} \{0,1\}^\ast\rightarrow\{0,1\}^n\\ \{0,1\}^n\rightarrow\{0,1\}^\ast \end{align}$$

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  • $\begingroup$ $\hskip -2.7em \color{black}{\Rule{2em}{1.8em}{2em}}$ $\endgroup$ – smileycreations15 Mar 22 at 16:56
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